The sum of the first six terms of an A.P is zero and the fourth term is 2. Findthe sum of its first 30 terms.

Given: T ₄ = 0

S ₆ = T ₁ + T ₂ + T ₃ + T ₄ + T ₅ + T ₆ = 0
Let the sum of first 30 terms be S ₃₀

And, the sum of first six terms be S ₆ .

T ₄ = 2

⇒ a + (4 – 1)d = 2
⇒ a + 3d = 2 …………..(i)

We know that,

Also,

S ₆ = 0

⇒ 2a + 5d = 0 …………..(ii)

(i)× 2,

a + 3d = 2

2a + 6d = 4 …………………..(iii)

(iii) – (ii),

(2a + 6d) – (2a + 5d) = 4 – 0

⇒ 2a + 6d – 2a – 5d = 4

⇒ d = 4
Substituting the value of d = 4 in (i),
a + 3 × (4) = 2

⇒ a + 12 = 2
⇒ a = 2 - 12 = - 10

∴ T ₃₀ = a + 29d
= - 10 + 29 × (4)
= – 10 + 116

= 106
∴ Sum to first 30 terms,

⇒ S ₃₀ = 15 × 96
⇒ S ₃₀ = 1140.